BIO97 EXAMS

in peas, the yellow allele is dominant over the green allele. A plant with yellow peas was crossed to a plant with green peas. The resulting plants were 50% yellow and 50% green. What are the genotypes of the parents in this cross?

Yy X yy

- the genotype of the green plant must be yy since the green allele is recessive to the yellow allele. The yellow plant can be either Yy or YY, but must be Yy in this case otherwise there would be no green offspring.

LEARNING OBJECTIVE: PREDICT THE PHENOTYPE OF AN ORGANISM GIVEN ITS GENOTYPE AND VISE VERSA; USE PUNNET SQUARES AND PROBABILITY RULES TO PREDICT CHARACTERISTICS OF OFFSPRING IN MONOHYBRID AND DIHYBRID CROSSES

Only sixty-one of the sixty-four codons specify an amino acid. In what process do the other three codons function?

Termination of translation (UAU, UAG, UGA)

- There are three stop codons which indicate to the ribosome to stop translation

LEARNING OBJECTIVE: EXPLAIN THE MOLECULAR BASIS OF TRANSLATION

Duchenne muscular dystrophy in humans is an x-linked recessive trait. If a heterozygous female has children with an affected male, then what is the probability of the children having muscular dystrophy?

Half of the children will have muscular dystrophy, regardless of sex.

- Based on the description, the female i Cc and the male is c-. Whether a child is affected will depend on which allele they inherit from their mother. Therefore, 1/2 the children of either sex will be affected (cc or c-)

LEARNING OBJECTIVE: EXPLAIN HOW SEX-LINKED GENES ARE INHERITED DIFFERENTLY THAT AUTOSOMAL

The DNA sequence below encodes the first five amino acids of a large protein

5'ATGTTAGGATATCAG3'

3'TACAATCCTATAGTC3'

Which strand is the coding strand?

The top strand

- Coding strand must have ATG start codon

LEARNING OBJECTIVE: TRANSCRIBE A DNA SEQUENCE INTO AN mRNA TRANSCRIPT; TRANSLATE AN mRNA SEQUENCE INTO A PROTEIN SEQUENCE USING THE GENETIC CODE

The DNA sequence below encodes the first five amino acids of a large protein

5'ATGTTAGGATATCAG3'

3'TACAATCCTATAGTC3'

In many organisms RNA editing can occur. This is a process by which a cell can make change to RNA transcript after it has been transcribed. Based on the DNA sequence above, if an A nucleotide were added after the third nucleotide in the RNA transcript, what would the first two amino acids in the protein be?

Met-Ile

- The new transcript, after editing, would be 5'-AUGAUUA... so the first two amino acids would be Met-Ile.

LEARNING OBJECTIVES: TRANSCRIBE A DNA SEQUENCE INTO AN mRNA TRANSCRIPT; TRANSLATE AN mRNA SEQUENCE INTO A PROTEIN SEQUENCE USING THE GENETIC CODE

The TATA sequence is found only several nucleotides away from the start site of transcription. This most probably relates to which of the following?

The number of hydrogen bonds between A and T in DNA

- Bc A-T has fewer hydrogen bonds than G-C, the TATA box, which is completely comprised of A's and T's it will more easily melt than a DNA sequence with a lot of G's and C's. This probably helps the DNA melt to allow for transcription

LEARNING OBJECTIVE: DESCRIBE THE STRUCTURE AND FUNCTION OF DNA, FROM SINGLE MOLECULES TO CHROMOSOMES; EXPLAIN HOW EUKARYOTIC GENE EXPRESSION IS REGULATED

If a eukaryotic chromosome was composed of 20% adenine, how much cytosine should theoretically be present in the same chromosome?

30%

- In double stranded DNA, %A=%T and %G=%C, and all the percentages must add up to 100%. To 20%A=20%T and therefore 30%G=30%C.

LEARNING OBJECTIVE: DESCRIBE THE STRUCTURE AND FUNCTION OF DNA, FROM SINGLE MOLECULES TO CHROMOSOMES

Hemophilia is an x-linked recessive disorder that affects a protein required for blood clotting. The dominant and recessive allele for the hemophilia-related alleles are H and h. Albinism is an autosomal recessive condition that results from a mutation of the gene that produces an enzyme in the melanin synthesis pathway. The dominant and recessive alleles related to albinism are A and a.

What could be the genotype and phenotype of Charles' sister?

Hhaa; albinism

- From the pedigree, you can see that Charles' sister (II-4) has albinism but not hemophilia. Since she does not have hemophilia, her genotype w/respec to those alleles can be Hh or HH but not hh.

-LEARNING OBJECTIVES: USE PEDIGREES TO TRACK THE INHERITANCE OF AUTOSOMAL AND SEX-LINKED TRAITS AND TO PREDICT THE CHARACTERS OF OFFSPRING

Determine the probability that Clara and Charles have a boy w/ hemophilia

1/4

- the probability that Clara and Charles have a boy is 1/2. Since Clara is healthy but has a father w/ hemophilia, her father must be h- and she must be Hh. Since hemophilia is x-linked, the probability that a boy of Clara's has it depends on the likelihood that she passes down the h allele, which is 1/2. Since these two events are independent, we multiply them together to get 1/2*1/2= 1/4

LEARNING OBJECTIVES: USE PEDIGREES TO TRACK THE INHERITANCE OF AUTOSOMAL AND SEX-LINKED TRAITS AND TO PREDICT CHARACTERS OF OFFSPRING

Identify which of the following includes three possible components of an RNA nucleotide?

ribose, adenine, phosphate group

- RNA has ribose, not deoxyribose; RNA has uracil, not thymine; Both DNA and RNA have phosphate groups.

LEARNING OBJECTIVE: EXPLAIN THE MOLECULAR BASIS OF TRANSCRIPTION

In peas, axial (A) flower position is dominant to terminal (a) and tall (L) is dominant to short (l). If a plant that is heterozygous for both traits is allowed to self-fertilize, how many of the offspring would show the dominant phenotype for both traits?

9/16

- The parents are AaLl. If you do the punnet square for flower position, you find 3/4 of the offspring will show the axial phenotype. Likewise, 3/4 of the offspring will show the tall phenotype. Since these traits are independent, the probability of both phenotypes being observed is 3/4*3/4=9/16

LEARNING OBJECTIVE: USE PUNNET SQUARES AND PROBABILITY RULES TO PREDICT CHARACTERISTICS OF OFFSPRING IN MONOHYBRID AND DIHYBRID CROSSES

If A,B, and C represent G1,S, and G2 respectively, then which phase of the cell cycle would you predict to last the longest in duration? Use the cells without treatment graph to answer this question

G1

- You would predict that the phase that is longest in duration is the one w the most cells in it. Therefore, for untreated (control) cells, the most cells are in bin A which corresponds to G1.

LEARNING OBJECTIVE: EXPLAINS WHAT HAPPENS IN EACH STEP OF THE CELL CYCLE INCLUDING MITOSIS

At which stage of the cell cycle did inhibitor93 stop progression through the cell cycle?

S

- you would predict that cells would accumulate in the phase at which the inhibitor stopped progression. Therefore, from the graph of the test group, you wold predict the drug stopped the cell cycle in bin B which corresponds to S phase.

LEARNING OBJECTIVE: EXPLAIN WHAT HAPPENS IN EACH STEOP OF THE CELL CYCLE INCLUDING MITOSIS

If an inhibitor was given to cells going through meiosis that blocked separation of chromosomes during meiosis II, what is the most immediate predicted effect?

Sister chromatids fail to separate

- in meiosis II sister chromatids pair and subsequently separate

LEARNING OBJECTIVE: EXPLAIN WHAT HAPPENS IN EACH STEP OF MEIOSIS AND HOW IT REDUCES CHROMOSOME NUMBER

It became apparent to Watson and Crick after completion of their model that the DNA molecule could carry a vast amount of hereditary information in which of the following?

Sequences of bases

- There are four possible bases in a DNA structure and these are the letters that can be used to carry genetic info

LEARNING OBJECTIVES: DESCRIBE THE STRUCTURE AND FUNCTION OF DNA FROM SINGLE MOLECULES TO CHROMOSOMES

DNA replication in a cell, PCR, and dideoxy sequencing have many common feature, Which is not a common feature of the three processes?

Many copies of the DNA template are required

- Many copies of a DNA template are absolutely required for dideoxysequencing, but a single copy of the DNA template is all that is needed for DNA replication in the cell

LEARNING OBJECTIVE: CONTRAST DNA REPLICATION, PCR, AND SEQUENCING

You are running a PCR and start with 4 copies of DNA. You need to have at least 40 copies of the DNA at the end. What is the minimum number of PCR cycles that you need to get 40 copies of DNA?

4

- you need at least 4 cycles of PCR. You double DNA after each cycle, so 1 cycle will give you 8 copies, 2 will give you 16, 3 will give you 32, and 4 will give you 64.

LEARNING OBJECTIVE: OUTLINE THE PROCESS OF PCR

These are the promoter sequences of 4 bacterial genes. The consensus sequence for the -10 region is TATAAT. Based on only the -10 sequence, which gene do you predict to have the highest expression?

Gene 1

- Gene 1 has a -10 sequence with only one difference within the consensus sequence. Gene 2 has 3 differences, Gene 3 has 2, Gene 4 has 2. Therefore, you can predict that Gene 1's promoter will recruit RNA polymerase the best and drive the most expression

LEARNING OBJECTIVE: PREDICT CONSEQUENCES OF MUTATIONS IN PROMOTERS

What is the best explanation for the observation that a human can produce more than 100,000 distinct proteins but the human genome only has about 20,000 genes?

Alternative splicing allows for one gene to encode many proteins

- Although polycistronic transcripts allow a single gene to code for multiple proteins, they are not found in humans

LEARNING OBJECTIVE: LIST THE SIMILARITIES AND DIFFERENCES BETWEEN EUKARYOTIC AND PROKARYOTIC TRANSLATION

Below is a diagram of cellular process. Which of the following statements is true?

The F pointer is nearest the N terminal of the protein

- F must be near the N terminal of the protein since it is the part farthest from the ribosome and proteins are synthesized from N to C terminal.

The C pointer is nearest the 5' end, not the 3' end. Since mRNAs are synthesized 5' to 3', the C pointer which is farthest away from the site of transcription must be at the 5' end.

The process must be in bacteria because the transcription and translation are shown; and the molecule closest to B is RNA polymerase

LEARNING OBJECTIVE: DESCRIBE THE PROCESS OF TRANSCRIPTION; DESCRIBE TRANSLATION INITIATION, ELONGATION AND TERMINATION; LIST THE SIMILARITIES AND DIFFERENCES BETWEEN EUKARYOTIC AND PROKARYOTIC TRANSLATION

A geneticist is conducting a genetic screen to study the Hedgehog signaling pathway and isolates a mutant fruit fly w small wings due to a null mutation, caused by a frameshift, in an important signaling protein named Patched. The geneticist subjects this mutant fly to further mutagenesis to identify revertants. Which of the following best describes a second-site reversion that is likely to restore the wild-type wing size?

A hypermorphic mutation in a protein that can carry out the same function as Patched.

- A hypermorphic mutation in a protein w the same function as Patched would be an example of a second-site reversion. A missense mutation in an unrelated pathway wouldn't restore WT phenotype, and a deletion that restores the reading frame in the mutated Patched protein is an example of an intragenic, not a second site reversion. A loss-of-function mutation in a protein with the same function as Patched would make the phenotype worse, not more wild-type

LEARNING OBJECTIVE: DISTINGUISH BETWEEN FORWARD AND REVERSE MUTATIONS

DNAse I is an enzyme that can randomly cut RNA when the chromatin is open, but not when it is closed. Regions that can be cut are called " DNAse I hypersensitive regions" Which of the following regions do you expect to be DNAse I hypersensitive?

The string part of beads on a string

- Barr bodies, heterochromatin, and repressed enhancers would all be inaccessible to DNAse I. "Beads on a string refers to the 10nm fiber of DNA where the beads are nucleosomes and the string is the linker DNA. The linker DNA should easily be cut by DNAse I

LEARNING OBJECTIVE: EXPLAIN NUCLEOSOMAL STRUCTURE AND CHROMATIN ORGANIZATION IN DIFFERENT PARTS OF THE CELL CYCLE

A male mouse with brown fur is mated to two different female mice with black fur. Black female 1 produces 9 black and 7 brown pups. Black female 2 produces 14 black pups. Black fur is _______ to brown fur and black female 1 is ______.

Dominant; heterozygous

- Given these results, black fur must be dominant to brown fur, and the black female 1 must be Bb, giving rise to roughly 1/2 black Aa and 1/2 brown aa offspring.

LEARNING OBJECTIVE: PREDICT PHENOTYPIC/GENOTYPIC RATIONS IN CROSSES OF PARENTS DIFFERENT AT A SINGLE GENE LOCUS OR TWO INDEPENDENTLY ASSORTING LOCI

A plant has the genotype AABbCcDd. These genes are involved in flower shape and exhibit epistasis. How many different types of gametes can result from this plant?

8

- The fact that their is epistasis is irrelevant, since we are asking about genotypes not phenotypes. This plant can produce 222 unique gametes (2 possible b alleles, 2 possible c alleles, and 2 possible d alleles).

LEARNING OBJECTIVE: USE THE PUNNET SQUARE TO DETERMINE GENOTYPE AND PHENOTYPE AND WRITE OUT THE GAMETE FROM EACH PARENT

A couple wishing to have a child consults a genetic counselor. The potential mother III-4 has the X-linked recessive trait, hemophilia A running in her family, as shown in the pedigree below. What is the probability that III-4 and III-5 have a child w/ hemophilia?

1/8

- Person III-4's mom is person II-1. She must be a carrier XAXa, since her father was affected. Person III-4's dad is a hemizygote, carrying the dominant allele XAY, since he is unaffected. Therefore there is a 1/2 probability that III-4 is a carrier (XAXa). The potential dad III-5 is XAY, since he is unaffected. Doing the punnet square, you can see that none of the daughters could be affected, since they will get a normal gene copy from their father, but half their sons would be affected, if they inherit the mutant allele from the mom. Therefore, the probability of an affected child is 1/2 (prob mom is a carrier)1/2(prob child is a boy) 1/2(prob boy has hemophilia)

LEARNING OBJECTIVE: PREDICT HOW X-LINKAGE CAN AFFECT PHENOTYPES AND INHERITANCE PATTERNS; APPLY THE PRODUCT, SUM, AND CONDITIONAL PROBABILITY RULES

Examine the accompanying pedigree and determine the inheritance pattern. Assume that penetrance is 100%

This trait could be autosomal recessive or x-linked recessive

- Dominant won't work since you cant have two unaffected parents w/ affected children. If it is autosomal recessive, both parents would be heterozygotes. If it is x-linked recessive, the mom is a heterozygote and the dad is a dominant hemizygote.

LEARNING OBJECTIVE: DRAW A PEDIGREE AND PREDICT INHERITANCE PATTERNS, GENOTYPES, AND PHENOTYPES

You are observing a cell w/ genotype AaBb. Genes A and B are on different chromosomes. Which of the following diagrams shows an appropriate alignment of chromosomes during metaphase I of meiosis

Cell B

- Cell B shows the the proper alignment. During metaphase 1, chromosome consist of two chromatids, and homologous chromosomes pairs are in complex and lie on the metaphase plate with one chromosome on each side, making B the right configuration. Cell A is not a possible configuration for metaphase I because the homologous chromosomes are not paired, Cell E shows metaphase in mitosis. Cell D is wrong because homologous chromosomes have the same genes, not different ones, and Cell C is not realistic, since it shows chromosomes with chromatids with different genes

LEARNING OBJECTIVES: COMPARE & CONTRAST CHROMOSOMAL BEHAVIOR IN MITOSIS AND MEIOSIS

You are studying a pathway that controls fruit fly wing size and have isolated two homozygous recessive mutants with small wings. When crossed, they yield offspring with normal wing size. Which of the following statements is true?

The two mutants have mutations in different genes

- This is n example when two mutants complement each other; therefore the mutations are in different genes. Statements A and D are false. Since two genes are involved, you can infer that your mutants are aaBB and AAbb from the info in the question. Therefore, you can predict the genotypic ratio resulting from a cross of the offspring (AaBb*AaBb)

LEARNING OBJECTIVE: DESCRIBE EPISTATIC VARIATIONS FROM CLASSIC MENDELIAN GENETICS; IDENTIFY COMPLEMENTATION GROUPS AND RELATE COMPLEMENTATION GROUPS TO GENES

Parakeet feather color is controlled by two genes. Yellow pigment is synthesized by the gene product of the dominant allele Y and blue pigment is synthesized by the gene product of the dominant allele B. B and Y are independently assorting. Parakeets that can't produce either pigment are white, those that can only produce blue pigment are blue, those that can only produce yellow pigment are yellow, and those that produce both are green. You cross a green and a yellow parakeet and get green, blue, yellow, and white birds. What are the genotypes of the parents?

BbYy*bbYy

- based on the description, the green parent must be B_Y_ and the yellow parent must be bbY_. since all colors of offspring are observed, both parents must be heterozygous for the dominant traits they show

LEARNING OBJECTIVES: DESCRIBE EPISTATIC VARIATIONS FROM CLASSIC MENDELIAN GENETICS; EXPLAIN HOW THESE RESULTS IN ALTERED PHENOTYPIC RATIONS IN F1 AND F2 GENERATIONS

Parakeet feather color is controlled by two genes. Yellow pigment is synthesized by the gene product of the dominant allele Y and blue pigment is synthesized by the gene product of the dominant allele B. B and Y are independently assorting. Parakeets that can't produce either pigment are white, those that can only produce blue pigment are blue, those that can only produce yellow pigment are yellow, and those that produce both are green. The result of the cross between the green and yellow parakeet above is 12 green, 4 blue, 13 yellow, and 3 white. Calculate the chi-square value for these data, using the description of inheritance described above

0.33

- Using the description above, you can do a punnett square and figure out that the predicted phenotypic ratio is 3:1:3:1 for green:blue:yellow:white. The expected numbers of offspring are 12:4:12:4. The chi-square= (12-12)^2/12 + (4-4)^2/4 + (13-12)^2/12 + (3-4)^2/4 = 0.33

LEARNING OBJECTIVE: TEST PROPOSED INHERITANCE PATTERNS WITH A CHI-SQUARE TEST

Parakeet feather color is controlled by two genes. Yellow pigment is synthesized by the gene product of the dominant allele Y and blue pigment is synthesized by the gene product of the dominant allele B. B and Y are independently assorting. Parakeets that can't produce either pigment are white, those that can only produce blue pigment are blue, those that can only produce yellow pigment are yellow, and those that produce both are green. How many degrees of freedom are associated with the chi-square test above?

3

- there are four categories, so df=4-1=3

LEARNING OBJECTIVE: TEST PROPOSED INHERITANCE PATTERNS WITH A CHI-SQUARED TEST

Which of the following statements is true of a heteroplastic gene?

A heteroplastic gene is found in multiple allelic forms in DNA molecules that make up the genomes of organelles in the cells of an organism

- definition of heteroplasmy

LEARNING OBJECTIVE: DESCRIBE THE ORGANIZATION, NUMBER, AND REPLICATION OF MITOCHONDRIAL GENOMES

A scientist has carried out a genetic screen to identify genes responsible for heart development, using zebrafish as a model organism. Which of the following is false?

The genes identified in the screen are unlikely to be important in human heart development

- Most developmental patterning genes are conserved between model organisms and humans

LEARNING OBJECTIVES: RATIONALIZE WHY MODEL ORGANISMS ARE NEEDED TO STUDY DEVELOPMENTAL BIOLOGY, AMONG OTHER BIOLOGICAL PROCESSES; DESCRIBE A GENETIC SCREEN THAT CAN HELP BIOLOGISTS UNDERSTAND THE PROCESS OF DEVELOPMENT

In 1718 the tsar of russia and his wife, the tsarina, and several of their children were assassinated and buried together. Later, several women claimed that they were anastasia, one of the tsars and tsarinas missing daughters. Scientists have extracted DNA from the skeletal remains. At a minimum, whose mtDNA would need to be analyzed to discern which, if any of the claimants was Anastasia?

The tsarina and the claimants

- Since this is mtDNA it follows a maternal inheritance pattern. All children of the tsarina would have the same mtDNA as her so you only need to examine the mtDNA of the tsarina and the claimants.

LEARNING OBJECTIVE: PREDICT THE INHERITANCE PATTERNS OF GENES ENCODED ON THE ORGANELLAR GENOMES AND WHY MITOCHONDRIAL INHERITANCE IS USEFUL FOR GENEALOGICAL STUDIES

You perform a three point cross using genes X,Y, and Z which are located in that order (XYZ) on the chromosome. Following convention, uppercase alleles are dominant and lowercase alleles are recessive. In the F0 generation, you use one x Y Z individual from a true breeding population, and one X y z individual from another true-breeding population. Which F2 genotype requires a double crossover event?

x/x y/y Z/z

- three point crosses

An analysis of Senator Warren's DNA was recently released. Five segments of Senator Warren's genome were identified as having Native American origin. The report states that " the total length of the 5 genetic segments identified as having Native American ancestry is 25.6 centiMorgans, and they span approximately 12,300,000 DNA bases."

What is the frequency of crossovers in the Native American genome region in Senator Warren's genome?

51.2%

-by definition of map distance: crossover frequency= twice the map distance

The size of the human genome is approximately 6,200cM. How many crossovers are there one average during human meiosis?

124

- number of crossovers= [(cM*2)/100]

The size of the human genome is approx. 3,200,000,000 base pairs. Comparing the ratio of map distance to physical distance, what could you expect of the Native American region in Senator Warren's DNA?

It may be slightly gene rich

- The map distance ratioed to physical distance in the Native American region is roughly 2.010^-6 cM/bp which is slightly higher than that in the overall genome, which is roughly 1.910^-6 cM. Gene rich regions are associated w/ euchromatin, more conducive to crossing over

Assume for this question and the next that Senator Warren's mother has Native American ancestry, but her father does not. What is true of Senator Warren's children?

Some of them may inherit DNA with Native American ancestry

- Senator Warren must be heterozygous in the region of interest. Therefore, each child has an independent 50% probability of inheriting the version with Native American ancestry.

Consider a line of Senator Warren's descendants that goes for 10 generations. Ignoring the possibility of crossovers, what is the probability that the 10th descendant inherits Senator Warren's largest segment of Native American DNA?

0.1%

- At each generation there is a 50% probability that the segment of interest will be transmitted. The probability that this happens 10 times in a row is 0.5^10=0.1%

The report states " the largest segment identified as having Native American ancestry is on chromosome 10. This segment is 13.4 cM in genetic length and spans approx. 4,700,000 DNA bases." If one of Senator Warren's children inherits least some of the largest DNA segment with Native American ancestry and ignoring double or multiple crossovers or the possibility that the child's father carries Native American DNA, what is roughly speaking the probability that this child inherits all 4,700,00 bases in the segment?

73.2%

- crossover frequency= 13.4*2=26.8%

Neglecting the possibility of double crossovers, for all of the region to be inherited there needs to be 0 crossovers occurring in the region - which will be the case 100%-26.8%= 73.2% of the time

Consider a line of Senator Warren's descendants who all inherit the largest DNA segment with Native American ancestry and who do not marry with individuals carrying Native American DNA. What is roughly speaking the probability that the 10th generation the segment still has its original length of 4,700,000 bases?

4.4%

- The probability that in each of the generations in the segment is not broken up is 0.732^10, since what happens at each generation is independent of what happened at the previous generation (0.736 is the probability computed in the previous question that no crossover takes place within the segment

You want to compute map distance between three genes C, Z, and X that each have two alleles, one dominant and one recessive (C, c, X, x, Z, z). You derive data following the test cross principle discussed in class. The information that you recovered is as follows:

What is the observed recombination frequency between x and c?

15.4%

- (15 + 10 + 89 + 91)/(1330)

The gene order is Z X C as determined by comparing a double recombinant with the parental types Z x c and z C X

What is the observed recombination frequency between z and c?

23.9%

- (89 + 91 + 68 + 71)/(1330)

What is the best estimate of the map distance between x and c?

15.4 cM

- using recombination frequency is the best we can do; computed above

What is the best estimate of the map distance between z and c?

27.7 cM

- This is the longer interval; sum map distances for z-x and x-c

12.3 + 15.4 = 27.7 cM

What is the best estimate of the map distance between x and z?

12.3 cM

- using recombination frequency is the best we can do; (15 + 10 + 68 + 71)/(1330)

Which of the following would you expect would not be a feature by which the presence of a particular plasmid could be selected?

Sensitivity to an antibiotic

- it is difficult to select for sensitivity to an antibiotic ( the converse, selecting for resistance)

Given three genes with dominant alleles A, B, and C and recessive alleles a, b, and c, which of the following represents a sensible mating for the second cross of a three-point-cross experiment?

Both A and D are correct, ABC/abc abc/abc and aBc/AbC abc/abc

- In the other answers, none of the parents is homozygous recessive for all genes; therefore, it would not be possible to infer parent gamete genotype from F2 phenotype

Eye color (E) and hair color (H) are imagined to be linked. If an individual with genotype HHee breeds with an individual with genotype hhEE, which of the following sets of gametes will be the most common in their progeny?

He and hE

- these are the parental types, which are more frequent than HE and he since the genes are linked

Which of the following statements is false?

a. genes on the same chromosome may or may not be linked

b. genes that are completely linked will not assort independently

c. genes on the same chromosome are systemic

d. genes on the same chromosome will assort independently if they are in a genome region where crossovers cannot take place

d. genes on the same chromosome will assort independently if they are in a genome region where crossovers cannot take place

- if no crossovers can take place in-between two genes, the genes will show complete linkage - which is the opposite of assorting independently